尝试从作为服务启动的python脚本启动程序

以python脚本的形式创建了一个服务，其中带有特定播放列表的VLC应该通过命令启动。 服务列表：

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs$
<plist version="1.0">
<dict>
    <key>Label</key>
    <string>com.bioxakep.biobot.plist</string>
    <key>Program</key>
    <string>/Users/BioMac/Documents/Scripts/newHome.py</string>
    <key>KeepAlive</key>
    <true/>
    <key>StandardOutPath</key>
    <string>/tmp/biobot.out</string>
    <key>StandardErrorPath</key>
    <string>/tmp/biobot.err</string>
</dict>
</plist>

VLC尝试启动并挂起，但在日志视图错误中：

LSOpenURLsWithRole () failed for the application /Aplplications/VLC.app with error -600 for the file /Users/BioMac/Desktop/Radio.m3u.
LSOpenURLsWithRole () failed for the application /Aplplications/VLC.app with error -10810 for the file /Users/BioMac/Desktop/Radio.m3u.

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请帮助我理解…

我认为您应该使用

Console.app

（使用Ctrl+Space显示）来发现有关挂起的更多信息

有关代码-10810的更多信息可在此处找到：

有关代码-600的更多信息，请访问：

LSOpenURLsWithRole () failed for the application /Aplplications/VLC.app with error -600 for the file /Users/BioMac/Desktop/Radio.m3u.
LSOpenURLsWithRole () failed for the application /Aplplications/VLC.app with error -10810 for the file /Users/BioMac/Desktop/Radio.m3u.